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The Six-fold Basis of Reality: The Proton Revisited

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DOI: 10.13140/RG.2.2.32939.41762

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Ian Beardsley

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! of 1 66

The Six-fold Basis of Reality: The Proton Revisited

Ian Beardsley

! of 2 66

Contents

Introduction………………………………………………………3

1.0 Inertia………………………………………………………….4

2.0 The Constant k……………………………………………..8

3.0 The Proton Radius…………………………………………12

4.0 Discussion…………………………………………………….15

5.0 Solar Eclipsing Moon……………………………….22

6.0 The Proton Charge…………………………………..27

7.0 The Solar Magnetic Field………………………….29

8.0 Conclusion……………………………………………..41

9.0 The Proton Revisited……………………………….43

Appendix 1………………………………………………………….63

Appendix 2………………………………………………………….64

! of 3 66

Introduction

Here I hypothesize an equation of state for the periodic table of the elements from Avogadro’s

number, and determine from that a constant k based on the idea that there is a constant for the

intermediary scale between protons and stars using the Chandrasekhar limit for white dwarf

stars. After constructing a six-fold theory for inertia, or mass, based on the idea that a proton is

a three dimensional cross-section of four dimensional hypersphere, I successfully predict the

radius of a proton. In the course of this we find the second is a Natural unit of time both in terms

of the proton, and gravitational constant G and Planck’s constant h, and in terms of the Earth/

Moon orbital mechanics. I show Avogadro’s number is a natural unit as well, both for the proton

and atom, the equation of state, or constant k, and for the macroscale of Moon/Earth/Sun. We

also see that this is a consequence of six-fold symmetry based on carbon the core element of life,

and hydrocarbons, the skeleton’s or biological life chemistry. We find as well we can determine

the charge of a proton with our constant k in terms of the six-fold, as well as the function of the

solar magnetic field at Earth orbit.

I really think Nature employs 6-fold symmetry because I think it is the most dynamic being the

product of the two smallest prime numbers 2 and 3. Those are the smallest factors down to

which anything can be reduced, so that is why I think 6-fold is the basis of Nature, that the basis

has to be reduced to the smallest factors. As such we have 2x3=6, 3x3=9, 2x9=18 and 3x6=18.

The periodic table of elements is periodic over 18 groups, which means when you count to 18 the

elements start over making groups where they have similar properties falling into the same

groups. These groups are determined by electron configurations, for instance carbon is in group

4 which means since it wants to have noble gas electron configuration, it gains 4 electrons to be

the same as the noble gas in group 18 of the same period. 18-14=4 meaning it ionizes as C4- so it

has 4 electrons to combine with 4 hydrogen atoms that are each H+ or so we can have two H

atoms that can combine with one oxygen atom which is O2- because oxygen is in group 16

meaning 18-16=2, and so forth making life possible.

Here in this version of the paper we revisit the proton and go into an explanation of the Van Der

Waal Radius from which our equation for the radius of a proton was derived. We also explain

how the Chandrasekhar limit was derived, but don’t derive it and leave that to the textbooks in

Astrophysics that all derive it in their chapters on stellar dynamics. The Chandrasekhar limit is

used here for the derivation of our intermediate mass, our constant k, which determines our

derivation of the unit of a second of time, and the charge of a proton, as well as the radius of a

proton. We find in this second look at the proton we can hone our equation for the radius of a

proton for more accuracy and what might be a more dynamic expression than the original

equation. We then look at the geometry that follows from it.!

! of 4 66

1.0 Inertia

We now need to show that the fundamental particles that build reality are based on sixfold

symmetry for their mass, size, and charge and we want to do it in terms of gravity on the

macroscale, thus it has to use G the universal constant of gravitation and Planck’s constant h,

that quantizes energy on the microscale. I find I can do this as such:

The natural constants measure the properties of space and time. We can list some of them here:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

Gravity is a property of space measured by the universal constant of gravity, G:

Equation 1.0

Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons

and neutrons) we will suggest given by:

Equation 1.2

Which describes mass per meter over time, which is:

Equation 1.3

It must be adjusted by the fine structure constant . It is my guess the factor should be

which is 18,769.:

Equation 1.4

Because the fine structure constant squared is the ratio of the potential energy of an electron in

the first circular orbit to the energy given by the mass of an electron in the Bohr model times the

speed of light squared, that is it represents the ground state. It is

: 1.67262 × 10

−27

h:6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G:6.67408 × 10

−11

c:299,792, 459m /s

α : 1/137

= q

= 1.6022E − 19coulombs

= 8.988E9

G = 6.674E − 11N

= 6.674E − 11

⋅ kg

kg ⋅ s

1.82E − 16

kg ⋅ s

(1.82E − 16kg ⋅ s/m)(18,769) = 3.416E − 12kg ⋅ s/m

! of 5 66

Equation 1.5

Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a

hypersphere. Thus we consider the surface area of a proton, :

Equation 1.6

We take the square root to get meters:

Equation 1.7

We multiply that with the value we have in equation 1.4:

Equation 1.8

We find that the mass of a proton realizes if we divide this by six seconds:

Equation 1.9

That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:

Equation 1.10

Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the

skeletons of organic matter, the so-called hydrocarbons:

Equation 1.11

Equation 1.12

It seems the duration of a second is natural. If it is, since it was formed by a calendar based on

reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon

orbital mechanics. I find it is, that (See Appendix 1):

Equation 1.13

That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic

energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by

very small amounts over millions of years. The solar system has evolved towards this since the

explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive

the second in terms of something else. For now we have the mass of a proton as:

= 4π r

= 8.72E − 30m

= 2.953E − 15m

(2.953E − 15m)(3.416E − 12

kg ⋅ s

) = 1.009E − 26kg ⋅ s

1E − 26

6secon ds

= 1.67E − 27kg

1E − 26kg ⋅ s

1secon d

= 1E − 26kg = 6protons = 6m

1E − 26kg ⋅ s

6secon ds

1E − 26kg ⋅ s

1secon d

moon

earth

(EarthDay) ≈ 1second

! of 6 66

Equation 1.14

This way of looking at things is to say matter is that which has inertia. This means it resists

change in position with a force applied to it. The more of it, the more it resists a force. We

understand this from experience, but what is matter that it has inertia? In this analogy we are

suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As

such there has to be a normal vector holding the higher dimensional sphere in a lower

dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in

the plane there should be a force countering it proportional to the normal holding it in a lower

dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a

sphere. This counter force would be experienced as inertia. Our reasoning above in one equation

is:

Equation 1.15

That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:

Equation 1.16

18α

4πh

h4π r

= 6proton ⋅ secon ds = carbon(C )

h4π r

= 1proton ⋅ 6secon ds = hydrogen(H )

Fig. 1

! of 7 66

For time t greater than 6 seconds we have fractional protons. For t<6 we the have other

elements. This is another indication that a second is natural; Carbon and hydrogen are the

upper and lower limits for integer seconds.

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number. We

make a program that looks for close to whole number solutions so we can create a table of values

for problem solving (Program in Appendix 2):

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

So again, a very interesting thing here is looking at the values generated by the program, the

smallest integer value 1 second produces 6 protons (carbon) and the largest integer value 6

seconds produces one proton (hydrogen). Beyond six seconds you have fractional protons, and

the rest of the elements heavier than carbon are formed by fractional seconds. These are the

hydrocarbons the backbones of biological chemistry. So we have the six-fold symmetry in the

chemical skeletons of life.This is to say that Carbon equals one second produces the radius of a

proton.

Equation 1.17

Where the above is multiplied by one second to make the units work. Where t is exactly:

Equation 1.18 !

h4π r

4πh

= 8.288587 × 10

−16

m = 0.829f m

6α

h4πr

= 1.004996352seconds

! of 8 66

2.0 The Constant k

Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational

Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:

In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic

equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where

‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational

constant numerically, but neglecting any units.

I found I could eliminate the and at the same time get the six of the six-fold symmetry with

which I was working by considering Avogadro’s number .

I suggested there exists some k that serves as a constant that describes both the microcosmos

and macrocosmos from the proton, to the atoms, to planetary orbits. It is such that the square

root of it times the earth orbital velocity is 6, because we are guessing we are dealing with six-

fold symmetry as the basis of Nature. That is

Eq. 2.1

We have that k is

Eq. 2.2

This follows from what Warren Giordano noticed that

Eq. 2.3

Without the right units. I noticed since Avogadro’s number is that I

could introduce an equation of state for the periodic table of the elements:

Eq. 2.4

Eq 2.5

Let us say we were to consider Any Element say carbon . Then in general

We have

and

1 + α

6.02E 23atoms

k v

= 6

k =

8π

(1 + α) ⋅ N

𝔼

h(1 + α) ⋅ 10

= G

6.02 × 10

≈ 6 × 10

ℍ = 1

gram

atom

(1 + α)

H = 6.0003

⋅ s

𝔼

ℂ

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

ℂ =

6gram s

6protons

6(6E 23protons)

6gram s

! of 9 66

Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a

molar mass of 12. We have

12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus

And it follows that

We see in general since the atomic number Z is the number of protons in an atom that in general

this holds for all elements because

And

Therefore we always have:

is a variable, the number of protons in multiplied by Avogadro’s number.

Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:

Equation. 2.6

While we have masses characteristic of the microcosmos like protons, and masses characteristic

of the macrocosmos, like the upper limit for a star to become a white dwarf after she novas (The

Chandrasekhar limit) which is 1.44 solar masses — More mass than that and she will collapse —

we do not have a characteristic mass of the intermediary world where we exist, a truck weighs

several tons and tennis ball maybe around a hundred grams. To find that mass let us take the

geometric mean between the mass of a proton and the mass of 1.44 solar masses. We could take

the average, or the harmonic mean, but the geometric mean is the squaring of the proportions, it

is the side of a square with the area equal to the area of the rectangle with these proportions as

its sides. We have:

Equation. 2.7

We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .

We have the intermediary mass is:

Equation 2.8

ℂ = 6E 23

(1 + α)

⋅ N

ℂ = 6.0003kg

⋅

𝔼

Z ⋅ 6E 23protons

Z ⋅ gram s

𝔼 =

Z ⋅ gram s

Z ⋅ protons

⋅ 𝔼 = 6E 23

𝔼

(1 + α)

⋅ N

𝔼 ⋅ v

= 422.787kg

⊙

= 1.98847E30kg

= 1.67262E − 27kg

= (2.8634E 30)(1.67262 E − 27) = 69.205kg

! of 10 66

All we really need to do now is divide equation 2.6 by equation 2.8 and we get an even number

that is the six of our six-fold symmetry.

Equation. 2.9

The six of our six-fold symmetry.

We have something very interesting here. We have

This is:

Eq. 2.10

Where k is a constant, given

Eq. 2.11

It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf

will not have its gravity overcome the electron degeneracy pressure and collapse. The non-

relativistic equation is:

Let us approximate 0.77 with 3/4. Since we have our constant

Eq 2.12

Eq. 2.13

Then

Eq. 2.14

Since our constant k in terms the Chandrasekhar limit is

(1 + α)

⋅ N

𝔼 ⋅ v

= 6.1092 ≈ 6

69.205kg

6kg

= 6

k v

= 6

k =

773.5

M ≤ 0.77

ℏ

= 1.41 ⊙

k =

1 + α

⋅ N

𝔼

= Mm

ℏ

1/2

ℏ = h /2π

! of 11 66

Eq. 2.15

The fine structure constant squared is the ratio of the potential energy of an electron in the first

circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of

light squared, that is it represents the ground state. It is

Since

We are suggesting the earth orbit is the ground state for our planetary system. We suggest it

holds for any planetary system because k as we will see is a natural constant that solves many

physical problems on many levels, not just planetary systems but atomic systems and the

particles that make them up.

Let us now recall equations 1.15 and 1.16

While we have considered them to be proton-seconds because they are a mass divided by the

mass of a proton, we can consider these two masses to cancel and say they are equal to 1 second

and six seconds respectively. We have that carbon, which is to evaluate them at one second, is

the radius of a proton:

Eq. 2.16

This gives the radius of a proton is:

Eq 2.17.

Where . The experimental value of the proton radius is 0.833fm+/-0.014fm!

k =

8π

(1 + α) ⋅ N

𝔼

k v

= 6

h4π r

= 6proton ⋅ secon ds = carbon(C )

h4π r

= 1proton ⋅ 6secon ds = hydrogen(H )

h4π r

= 6secon ds

4πh

= 8.288587 × 10

−16

= 0.829f m

= 1secon d

! of 12 66

3.0 The Proton Radius

Thus we have the radius of a proton is given by carbon by evaluating at one second:

But to get that we have to multiply by one second and we need one second in terms of the atom

for a theory of the proton. I find we can do that…

Substitute for to get

We have now introduced the radius of a hydrogen atom . It seems we have to

divide by two which I think is because we are looking at packing of atoms. This radius of the

hydrogen atom is the Van Der Waals radius, which is the closest distance between two hydrogen

atoms noncovalently bound. It is 120 pm. Divide that by ck where 1/k is our constant

And we find

We have our equation for the radius of a proton

We only need to multiply it by to have the right units, and we get!

4πh

= 8.288587 × 10

−16

m = 0.829f m

h4π r

h4π

t =

2α

h4π

= 1.2E − 10m

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12seconds

3 2

⋅

Gπ

⋅

𝔼

= 1.12seconds

4πh

= 8.288587 × 10

−16

m = 0.829f m

t /ck = 1secon d

! of 13 66

Equation 3.1.

Then suggest we picked up 9/8 in approximations which is close to one anyway so we write

Equation 3.2.

We form constants:

And we have the Equation:

Equation 3.3

We can say that Avogadro’s number is not an arbitrary number because it is such that there are

twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add

to this section the equation of the mass of a proton.

If our equation is right and we put it into natural units then the product should be close to

one:

Let us start with the units with which we are working:

2 ⋅

4π

⋅

𝔼

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

(

)

(

)

(

)

ℍ

(

)

k =

2π

= 6.93E − 9kg

𝔼

= 1.99E − 34m

= k

𝔼

2π

⋅

𝔼

G =

kg ⋅ s

h = kg ⋅

! of 14 66

And convert these to proton-masses and proton-radii:

Now we find k in these units:

Thus we have:

c = m /s

G = 6.67408E − 11

kg ⋅ s

⋅ 1.67262E − 27kg(0.833E − 15m)

⋅ s = 193,131, 756

h = 6.62607E − 34kg ⋅

⋅

(0.833E − 15)

(1.67262E − 27kg)

= 5.71E 23

c =

(299,792, 459m /s)(1sec)

(0.833E − 15m)

= 3.6E 23

1.2E − 10m

0.833E − 15m

= 144,058

k =

2π

= 6.93E − 9kg

k =

(5.71E 23)(3.6E 23)

2π

(193131756)

= 4E18proton − masses

= k

𝔼

(4E18)(144058)

(6E 23)

5.76E 23

(6E 23)

= 0.96 ≈ 1

! of 15 66

4.0 Discussion

The crux is that

Is calculated with perihelion and at aphelion giving a near

perfect result for one second. And, . And,

. We have

If multiplied by one second, which gives

Where t is

Giving

Using

Where ck is m/s cancels with s/m, and k is

6α

h4π r

= 1.004996352seconds

moon

earth

(EarthDay) = 1.08seconds

moon

earth

= 30,290m /s

= 966m /s

Si xDays = 6(24)(60)(60) = 518400s

= carbon

4πh

= 8.288587 × 10

−16

= 0.829f m

2π

⋅

𝔼

t =

2α

h4π

= 8.26935E − 16m ≈ 0.827f m

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12seconds

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

! of 16 66

But as well we can get our one second from

We notice this is centered around the structure at the basis of life, the hydrocarbons

The k is derived from coupling these two :

Which holds for any element:

Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a

molar mass of 12. We have

12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus

And it follows that

We see in general since the atomic number Z is the number of protons in an atom that in general

this holds for all elements because

And

moon

earth

(EarthDay) = 1.08seconds

h4π r

= 6proton ⋅ secon ds = carbon(C )

h4π r

= 1proton ⋅ 6secon ds = hydrogen(H )

𝔼

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

ℂ =

6gram s

6protons

6(6E 23protons)

6gram s

ℂ = 6E 23

(1 + α)

⋅ N

ℂ = 6.0003kg

⋅

𝔼

Z ⋅ 6E 23protons

Z ⋅ gram s

! of 17 66

Therefore we always have:

is a variable, the number of protons in multiplied by Avogadro’s number. By using the

Chandrasekhar limit we define an intermediate mass from the mass of the Sun

We multiply this by 1.44 (1..44 solar masses is the Chandrasekhar limit) to get 2.8634E30kg.

The mass of a proton is . We have the intermediary mass is:

More explicitly using the Chandrasekhar limit:

Intermediary mass:

The constant k given explicitly:

The k is such that

And k using the approximation

We show the equation for one second in terms of the Natural constants on the next

page…!

𝔼 =

Z ⋅ gram s

Z ⋅ protons

⋅ 𝔼 = 6E 23

𝔼

⊙

= 1.98847E30kg

= 1.67262E − 27kg

= (2.8634E 30)(1.67262 E − 27) = 69.205kg

M ≤ 0.77

ℏ

= 1.41 ⊙

ℏ

1/2

k =

8π

(1 + α) ⋅ N

𝔼

k v

= 6

(1 + α) ≈ 1

k =

8π

⋅ N

𝔼

! of 18 66

! of 19 66

We discuss what interests me most on the next page, the two different equations for one

second that connect the macrocosmos (the Earth, Moon, and Sun) to the microcosmos,

the constants of the elements.!

! of 20 66

If we can predict the radius of a proton with

Equation 4.1.

Where is given by in terms of the Moon, Earth, and Sun:

Equation 4.2.

Or in terms of the Natural Physical Constants Using equations:

Equation 4.3.

Then to understand why the solar system is described in in terms of elemental states of

periodic table that we have formulated and in that this predicts the solar system as

something that can be described in a quantum mechanical manner, we equate the latter

two:

Eq. 4.4.

Eq. 4.5.

Which is

Eq 4.6

This seems to say something interesting to me. A mole does not just pertain to atoms

per gram of an element, or molecules per sample, but perhaps speaks of a number of

Moons, Planets, Stars, or even galaxies. Let’s analyze to try and determine how that

might work, we break equation 4.5 and 4.6 down into its factors (Next page):!

4πh

= 8.288587 × 10

−16

= 0.829f m

moon

earth

(Ear thDay) = 1.08seconds

16πα

⋅

3 2

𝔼

⋅

= 1.12seconds

moon

earth

(Ear thDay) =

16πα

⋅

3 2

𝔼

⋅

𝔼 =

16πα

⋅ 3 2 ⋅

earth

moon

⋅

Ear thDay

6.02E 23 =

16πα

⋅ 3 2 ⋅

earth

moon

⋅

Ear thDay

! of 21 66

We see that we might be speaking of 4.286E20 meters per kg mass perhaps in a Nebula

where stars form:

Equation 4.7.

We see the scaling is given by h/G by a factor of which is like comparing the small

(h) to the large (G):

Equation 4.8.

And finally this is given by a factor of the Earth to the Moon:

Equation 4.9.

There is also a fraction of:

Equation 4.10.

And, there is a frequency given by the Earth day, the time it takes to rotate:

We find something interesting. If these factors describe the Solar System the way

Avogadro’s number describes elements and their compounds, then it would make sense

because their product is Avogadro’s number and as such it applies not just to elements

and compounds but to planets and stars and the interstellar medium. And we see here

that

Describes the KE of Earth and Earth rotation period in terms of Avogadro’s number, the

mass of a proton, the radius of a hydrogen atom, and the mass of a proton, the latter two

being the primary constituents of the Universe, gas giant planets, and the interstellar

medium.!

16πα

(18769)1.2E − 10m

16π(1.672E − 27kg)

= 2.67875E19

1/m

(6.626E − 34)

(6.674E − 11)(1.672E − 27)

= 5,938kg

earth

moon

(5.972E 24kg)(30,290)

(7.34767E22kg)(966m /s)

= 79,912 . 5

3 2 = 4.2426

Earth Day

86,400

= 1.1574E − 5

c ycles

6.02E 23 =

16πα

⋅ 3 2 ⋅

earth

moon

⋅

Ear thDay

! of 22 66

5.0 Solar Eclipsing Moon

Essentially that the Moon perfectly eclipses the Sun as seen from the Earth means while

it is 400 times smaller than the Sun it is 400 times further from the Sun than it is from

the Earth. This determines its orbital velocity and mass, as well as that of the Earth and

the mass of the Sun. The orbital velocities of the Moon and the Earth are given by:

Equation 5.1. and

Equation 5.2.

The Moon perfectly eclipses the Sun because

Equation 5.3.

Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius,

and is the lunar radius. This gives:

Equation 5.4.

We have more explicitly:

Equation 5.5.

The rotational period of the moon is and the

orbital period of the Earth is is 27.3 which gives

⊙

⋅

⊙

⋅

1.989E30kg

5.972E 24

⋅

1.74E6m

6.957E8m

= 28.86

≈

⊙

⋅

⊙

= 28.6

⊙

⋅

⊙

⋅

1.74E6m

6.957E8

1.989E30kg

5.972E 24kg

⋅

1.496E11m

3.84748E8m

= 28.46

= 27.3days = 2,358, 720seconds

= 24hours = 86,400seconds

! of 23 66

Equation 5.6.

Thus while second is given by:

The Lunar month is given by

Equation 5.7.

We also found in terms of the proton, the second is given by

Which gives the radius of the proton as

Where is one second approximated by

Which gives the radius of a proton is

Which we derived using our constant k:

⊙

⋅

moon

earth

(Ear thDay) = 1.08seconds

⊙

⋅

⋅ Γ

6α

h4πr

= 1.004996352seconds

4πh

= 8.288587 × 10

−16

= 0.829f m

16πα

⋅

3 2

𝔼

⋅

= 1.12seconds

2 ⋅

4π

⋅

𝔼

k =

8π

⋅ N

𝔼

! of 24 66

The Earth day is then:

Equation 5.8.

But since we also have

Then we have:

Equation 5.9

Where

Which gives

Equation 5.10

Equation 5.11

R ⊙

⊙

⋅

⋅ T

moon

earth

(Ear thDay) = 1.08seconds

moon

earth

⋅

R ⊙

⊙

⋅

⋅ T

= 1.08s

6α

h4πr

= 1.004996352seconds

16πα

⋅

3 2

𝔼

⋅

= 1.12seconds

moon

earth

⋅

R ⊙

⊙

⋅

⋅ T

6α

h4πr

moon

earth

⋅

R ⊙

⊙

⋅

⋅ T

16πα

⋅

3 2

𝔼

⋅

! of 25 66

From equation 5.8 for the Earth day

Checking our work we see from equation 5.7 for the Earth month

One of the things that we have not done yet is equate the expression on the right of 5.10

with that on 5.11

Which is Avogadro’s number in terms of the size of a hydrogen atom, the mass of a

proton, the radius of a proton, and h and G:

Equation 5.12

Let see just what kind of prediction 5.10 brings

R ⊙

⊙

⋅

⋅ T

⊙

⋅

⋅ Γ

R ⊙

⊙

⋅

⊙

⋅

⋅ Γ

1 = 1

6α

h4πr

16πα

⋅

3 2

𝔼

⋅

𝔼 =

9 2R

8πr

4πG

moon

earth

⋅

R ⊙

⊙

⋅

⋅ T

= 1.08s

moon

earth

⋅

R ⊙

⊙

⋅

⋅ T

6α

h4πr

! of 26 66

=28.46

We convert that to days,

2.28568E6/{(60s)(60m)(24d)}=31.74556 days

The lunar month with respect to the Sun (sidereal month) is 27.3 days. And with respect

to the Earth (synodic month) is 29.53059 days. So we are in the right area. We have

suggested the second is a Natural unit, but we might say the month is as well. In a sense

we already knew this because we have always known the Moon perfectly eclipses the

Sun, and here we have shown that is determined by the mass of the Earth, the radius of

the Sun, its mass, and these determine the Moon’s orbital distance, size, and mass, as

well the Earth’s mass. The Moon also has a function; it allows for life on Earth because

its orbit holds the Earth’s inclination to its orbit which allows for the seasons.

6α

h4πr

⋅

earth

moon

⋅

⊙

⋅

earth

moon

(5.972E 24kg)(30,290)

(7.34767E22kg)(966m /s)

= 79,912 . 5

6α

h4πr

= 1.004996352seconds

⊙

⋅

1.74E6m

6.957E8

1.989E30m

5.972E 24m

⋅

1.496E11m

3.84748E8m

= (1.005s)(79,912 . 5)(28.46) = 2.28568E6seconds

! of 27 66

6.0 The Proton Charge

I construct the electric field as such (Fig. 2): is one component of . Events in

are through time with components and where c is the speed of light. The

electrons and the protons and in the presence of one another cross into giving

themselves acceleration in the and directions.

We have

We suggest

Equation 6.1

The position of changes in the space of and has travelled to a place in

time, as well. Since ct=meters there is a g such that . Thus since

ℝ

(x, y, z)

ℝ

(x, y, z)

−x



0 ct

0 0 ct

= (ct

− ct

)



··

x ∝ (ct

− ct

)



ℝ

(x, y, z)

t (0,ct

, ct

)

··

x = g(ct

− ct

)



g ⋅ m =

Fig. 2

! of 28 66

and that is . Thus g is frequency squared ( ) and we suggest it is

derived from the separation between the charges and . That there is some velocity v such

that . We call upon our equation for 1 second:

Letting v= from our equation for k

And we have

Equation 6.2

We get

Equation 6.3 !

g =

⋅

⋅ s =

f =

(

)

h 4π

= 1secon d

(

)

(c)

h 4π

⋅

(c)

h 4π

= k

⋅

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= 773.5

m = m

(

⋅

)

h 4π r

= q

h 4π r

(6.626E − 34)(4π)(0.833E − 15)

(6.674E − 11)(299,792, 459)

= 5.37E − 31

q = (0.033)

36(18769

)

773.5

(5.37E − 31) = 9.1435E − 19C

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71protons ≈ 6protons

! of 29 66

7.0 The Solar Magnetic Field

We model the formation of the solar system from a slowly rotating gas cloud, a nebula of

gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on

and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,

like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from

which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three

factors that determine its structure, the inward gravity, the pressure gradient outward which

balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The

flattened rotating disc is broken up into rings each that has a mass spread out over it from which

the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth

masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass

spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars

masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the

Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of

about 350 because it is mostly iron condensations with incomplete silicon condensations.

Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is

-1.5 so that the density distribution of the protoplanetary disc is:

Giving a mass

With pressure gradients playing the key role in the formation of solar system, less attention is

payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of

this aspect found something very interesting concerning it. He was Alfven (1942). At the time

people were suggesting instead of the solar system forming from a rotating nebula, rather the

sun came into existence not at the same time at the center of the disc, but rather passed through

clouds and captured material after already existing. He figured for the captured material its

inward component v, and density , at a distance r from the sun, had to conserve mass, which

required:

He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions

would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the

magnetic field becomes important. He considered for simplicity the solar magnetic field was

generated by a dipole moment , a vector quantity, and that a particle moving in the plane of

that vector with mass m and charge q, would have all of both the gravitational and magnetic

forces in that plane, so the problem becomes two-dimensional and required only the and , of

polar coordinates. The differential equations of its motion would be:

σ (r ) = σ

−3/2

= 3300

M =

∫

2π

∫

σ (r )r dr d θ

= 4π r

ρv

! of 30 66

Equation 7.1

And,

Equation 7.2

We can integrate equation 7.2 with the boundary condition that the angular momentum of the

particle is zero at large distances from the sun to get (Appendix 2):

Equation 7.3

And substitute it into equation equation 7.1 for to get

Equation 7.4

Which we can write

Equation 7.5

We then integrate this with respect to r with the boundary condition that at large r and get

Equation 7.6

He then notices there is another value for which . It is

Equation 7.7

This is interesting because it means the particle can never approach the Sun closer than this

value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this

as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the

magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact

that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the

region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen

and helium.

Certainly today we don’t see the planets as having formed from material gathered by the Sun in

its journey, but rather think the Sun and planets formed at the same time from a cloud that

collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through

clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing

clouds from falling into their star. But we can also apply his equation to our Sun today, for which

we know a great deal about its magnetic field, which also happens to be an important thing to

study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels

··

r = −

⊙

qμ

+ mr

d(r

θ )

= −

qμ

θ = −

∫

qμdr

qμ

··

r = −

⊙

= −

⊙

r = 0

2GM

⊙

−

r = 0

(

2GM

⊙

)

1/3

! of 31 66

with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic

field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict

solar maximums, so we have warning as to whether there will be a magnetic storm that will

knock out our electrical grid and internet, ahead of time.

During solar minimum the solar magnetic field has closed lines, that flow out one pole and into

the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so

that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The

total magnetic field of the Sun at the Earth, is all the components taken together, which are

, , and . The important component is , because it runs north-south, so it is

perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the

component that interacts with the Earth magnetosphere, and when it points southward, it will

connect with the Earth’s magnetosphere which points northward so the solar poles flow into the

Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain

down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t

always stay around the Sun itself, but the solar wind carries it through the solar system until it

collides with the interstellar medium reaching the heliopause. Thus the Sun creates the

Interplanetary Magnetic Field (IMF) which has a spiral shape because the Sun rotates once

about every 25 days. But the upshot is that at Earth we have

Moderate Magnetic Field: 10 nT

Strong Magnetic Field: 20 nT

Very Strong Magnetic Field: 30 nT

For our purposes we want to return to equation 55:

! of 32 66

And ask just what is , because in the time that Alfven was working we worked with magnetic

fields differently, aside from his equation uses a trick, which we still use today, and that is to

consider the magnetic field a dipole. To consider it like this is to say there are two monopoles

opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,

though they are predicted by some modern theories, they have never been found. The trick is in

that by treating the North magnetic pole and South magnetic pole as separate magnetic charges

is to treat them like we do electric charges, the charge of a proton and the charge of an electron,

which can be convenient for making computations, but don’t exist that we know of. So we will

solve equation 56 for , and see what its units are so we can understand what it represents and

we will let m be the mass of a proton and q the charge of a proton. We get:

Equation 7.8

We can write these units as, by taking Coulombs (C) equal to

Equation 7.9

This is units of force per current density, which makes sense because a flowing current creates a

force. We can also write it:

Equation 7.10

Which is energy per magnetic field strength in that the SI units of magnetic field strength is

amps per meter. This tells us:

Equation 7.11

Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and

make it a proton:

We have

, , ,

(

2GM

⊙

)

1/3

⊙

= μ =

⋅ kg

C ⋅ s

a m p ⋅ secon d s

⋅

a m ps

⋅

a m ps

μ =

En erg y

Magn et icFiel d Strength

H ⟶ H

+ e

−

= 1proton = 2.18E − 18 J

= 1.6E − 19C

= 1.67E − 27kg

G = 6.67408E − 11N

⊙

= 1.989E 30kg

! of 33 66

We want to look at the Sun as having a current flowing around its equator in a loop with its

radius

We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger

than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar

magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979E11m):

Equation 7.12

Equation 7.13

Or,…

Equation 7.14

The radius of a proton is 0.833E-15m. We have that r is:

Equation 7.15

⊙

= 6.957E8m

1AU

5.5362E12a mps

1.496E11m

= 37.0a m ps /m

μ =

Ioni z at ion Energ y

Magn et icFiel d Strength

2.18E − 18J

37.0A m ps /m

= 5.892E − 20

J ⋅ m

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

= 5.92 ≈ 6Pr oton Ra dii = carbon

! of 34 66

Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it

provided for by the Sun’s magnetic field.

! of 35 66

Let us write the computation as one equation, and verify it. We have

Where

=Ionization energy of Hydrogen

=3.47E-39 (correct)

Equation 7.16

(0.000034574)(3.47E-39)=1.19972E-43

1.19972E-43^(1/3)

=4.932E-15

So as you can see equation 7.15 is correct. It says that carbon, the basis of life is in the ratio of

the solar magnetic field and the solar gravitational field.!

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

⊙

1AU

5.5362E12a m ps

1.496E11m

= 37.0 a m ps /m

1AU = r

μ =

Ioni z at ion Energ y

Magn et icFiel d Strength

2.18E − 18J

37.0A m ps /m

= 5.892E − 20

J ⋅ m

⊙

1AU

⊙

= (5.982E − 20)

= 3.47E − 39

(IE

)

⊙

(2.18E − 18)

(12.56637E − 7)

(1.496E11)

4(5E − 3)

(6.957E8)

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.496E11m)

4(5E − 3T )

(6.957E8m)

1/3

! of 36 66

Thus we have a theoretical value for the radius of the proton:

Equation 7.17

And a theoretical value for its charge

Equation 7.18

Equation 7.19

And we have the radius of a proton in terms of the solar magnetic field at Earth

Equation 7.20

All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our

constant k

Equation 7.21

And making the approximation we can with these equations eliminate in equation

5.21 using equation 5.19 in which we can eliminate k with equation 2.22.

Equation 7.22

This has an accuracy of close to 88% because in equation 7.19 the charge of six protons is

predicted by the theory to be 5.72 protons, the rest of the equations are much more accurate, but

we seek to rectify that. We write this equation so we can have an equation that defines the solar

magnetic field by solving for .

To address the accuracy of the equation for the charge of a proton, equation 7.19, we ask what is

the culprit. We suggest it is . So we solve the equation for that to see by how much it is

off. It is:

Equation 7.23

We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%

accurate, but we want to do better.

2π

⋅

𝔼

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6proton s

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

k =

8π

(1 + α) ⋅ N

𝔼

1 + α = 1

31850496r

⋅

𝔼)

h 4π

(IE

)

⊙

(α

)/36

= q

⋅

⋅ k

h 4π r

! of 37 66

We now eliminate on the left in equation 7.22 with equation 7.17 to find the B field of the Sun

as described by the Earth orbit as the ground state:

Equation 7.24

Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives

a magnetic field strength of 4.73E-3 Teslas. Concerning equation 7.24

We want it to be more accurate. To do that we have to substitute for , the fraction

which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy

could be in as its experimental value has the largest errors. Other possibilities are the

is right and that as well this is due to the radius of a proton having large errors, or even that it is

supposed to be the factor 8, which would be good to consider because 8-fold symmetry is very

dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear

synthesis by stars.

Because we are looking at 6 protons we had , and because of six-fold symmetry we had

and because the charge we had because of the 1/3 root on the right. So taking

and leaving the that came from substituting for k, and the as

well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation

we have:

Equation 7.25

Where

Or we can factor out all the numbers on the right the 2 and 4 in equation 7.25 and write

Equation 7.26

In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the

radius of a hydrogen atom, its ionization energy, and the solar gravitational field, with the earth

orbit as the ground state.!

⊙

⋅

(IE

)

⊙

⋅

h 4π

= q

⋅

⋅ k

h 4π r

(α

)/36

(α

)/36

q = 6q

N = 6

= 279936

8 = 2

⊙

⋅

(IE

)

2GM

⊙

⋅

h 4π

N = 6

= 279936

⊙

⋅

(IE

)

⊙

⋅

h π

! of 38 66

Protons and neutrons pack in atoms in a way that can be seen from their electron clouds. Thus

since we can say the electron cloud of helium is spherical the nucleus packing is spherical. But

since the protons and and neutrons in the nucleus are moving (But only as far as their nuclear

wall) the packing is ordered but liquid.

I believe the reason for my finds of six-fold symmetry in both the radius of a proton and it its

charge can be explained by Buckminster Fuller’s vector equilibrium. It is the most

transformable straight-line geometry if you attach sticks with flexible corners, what Fuller calls a

flex-corner. See the following illustration.

Which means it zero-frequency for omni directional closest packing of spheres is 12 spheres,

which is carbon (six protons and six neutrons) which is the basis of life as we know it. Its

diameter is then six proton radii which we found provided for by the solar magnetic field. The

frequency is the number on any symmetrically concentric shell or layer and is given by

The more protons an element has the more neutrons it needs so the strong nuclear force in the

nucleus can overcome the mutual repulsion of protons, thus holding the atoms together.

10F

+ 2 = 10(1)

+ 2 = 12 = carbon

! of 39 66

Thus, to go over that of the solar magnetic field again, proposed by Hannes Alfven (1942), at the

time there was no known mechanism for it, but he suggested the relative velocity between a

neutral gas and a plasma has a critical velocity at which the gas starts to ionize and that the

atoms or molecules will not exceed this velocity until the gas becomes almost fully ionized. The

additional energy put into the system goes into ionizing the gas instead of the velocity of the

atoms, and is roughly independent of pressure and magnetic field. Critical ionization velocity

has been recognized in the laboratory for some time. It is given by equating the kinetic energy of

the atoms to the ionization potential:

Alfven found:

Gas cloud enters Solar System

A neutral atom falls into Sun due to gravity

Motion is random, collisions happen

Temperature rises

At a distance from the Sun gas will ionize

m=atomic weight

He found for a gas cloud with average voltage 12 volts. An average atomic weight of 7, the is at

Jupiter.

That is atoms fall in towards the Sun and ionize at which point the solar magnetic field pushes

them out to Jupiter orbit where a halo forms from which planets can form.

= eV

ion

= G

ion

= G

ion

= 13.5E10

ion

= volts

! of 40 66

Thus if we are to suggest, as we have, that solar magnetic field provides for life, whose skeletons

are the hydrocarbons, that this has something to due with the solar wind which is plasma and

we can consider it a plasma that can move in neutral gas thus being subject to Alfven’s critical

velocity. The plasma is mostly electrons, protons, alpha particles kinetic energies between 0.5

and 10 keV. Their are trace amount sof heavy ions of C, N, O, Ne Mg, Is, S, and Fe. We note the

C, N, O are the most abundant elements in life chemistry. The solar wind can reach velocities of

25o000-750000 m/s. Remember the critical velocity for hydrogen is 50,900 m/s.

! of 41 66

8.0 Conclusion

Thus in conclusion we gave carbon as one second determines the radius of a proton

Where one second is

That this is nearly perfectly one second makes the second dynamic because carbon is

And hydrogen is

Which is dynamic because we have integer seconds in terms hydrocarbons the skeletons of life

being the upper and lower limits for integers that produce protons in elements. That is one

second is the lower integer limit that determines whole protons and is carbon the core element

of life and anything greater than six seconds is fractional protons and is hydrogen. Interestingly

the second we find is described in terms of the constants and atomic dimensions and in terms of

something on a much larger scale, the Earth/Moon/Sun orbital parameters and that these in

turn determine the radius of a proton:

If multiplied by one second ( ) gives

Where is

4πh

= 8.288587 × 10

−16

m = 0.829f m

6α

h 4π r

= 1.004996352 secon d s

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydr ogen(H )

6α

h 4π r

= 1.004996352 secon d s

K E

moon

K E

earth

(Ear th Da y) = 1.08secon d s

4πh

= 8.288587 × 10

−16

= 0.829f m

2π

⋅

𝔼

! of 42 66

Giving

We see our constant k, determines the radius of a proton on the microscale and the Earth orbit

in terms of the six-fold

Where is the Earth orbital velocity. We see that k is

Where is an equation of state for the periodic table of the elements and is equal to

Avogadro’s number, making it a Natural number, not just an arbitrary large number for dealing

with large amounts of atoms. We see k is determined by our idea of an intermediary mass

between protons and the stars using the Chandrasekhar limit for white dwarf stars. We see we

can use our constant k to determine the charge of a proton in terms of our six-fold:

And that we can do the same with the solar magnetic field:

Which is to suggest the solar magnetic field perhaps balances with its gravity to make six proton

radii functional at Earth orbit. We see this in terms of Buckminster Fuller’s vector equilibrium:

t =

2α

h 4π

= 8.26935E − 16m ≈ 0.827f m

k v

= 6

k =

8π

(1 + α) ⋅ N

𝔼

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6proton s

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

! of 43 66

9.0 The Radius of a Proton Revisited

In section 3.0 that starts on page 12, we noted to determine the radius of a proton we needed

an equation for the unit of one second in terms of the Natural constants. We found the radius

of a proton was"

If we multiplied through by one second. We are guessing one second is a natural unit because we

found it is in the orbital mechanics of the Earth and Moon and indeed we measure time by the

periods of their motions. We found

If it is calculated with orbital velocities at perihelion and at aphelion

which gives a near perfect result for one second and still close in the average

orbital velocities, which have little variation because the orbits of these bodies are nearly circular

(have low eccentricities).

Our value for one second in terms of the Natural constants uses the radius of a hydrogen atom

as given by the Van der Waals radius, the hydrogen atom being the simplest element with just

one proton which is the basic building block of the periodic table of the elements. It makes sense

in our theory here to use the Van der Waals Radius, because our theory is holistic; it determines

its parts from the whole, and the whole from the parts. So the proton is a part of the hydrogen

atom, the most fundamental part, and indeed we can call a hydrogen atom a proton, which we

actually do in chemistry, we call it a positive hydrogen ion, because stripped of its electron the

hydrogen ion is simply a proton.

Our theory is that the microcosmos is related to the macrocosmos and is a function of it as much

as the macrocosmos is a function of the microcosmos. But there are many levels that constitute a

cosmos; the proton theory is nuclear chemistry or atomic physics, but the Van der Waals radius

is chemistry, chemistry; the elements made from the atomic particles. So these are two levels.

Another level up from the elements are the compounds, and from there we go to the planets,

which takes us to the stars, and from the stars we go to galaxies, from there to clusters of

galaxies, from here we can go to the universe and from there to the multiverse.

And indeed we are bridging the different layers of the cosmos, we are doing it with our constant

k. We have earth at the ground state for sixfold symmetry:

Where k is

And k determines the time of one second in

4πh

= 8.288587 × 10

−16

m = 0.829f m

K E

moon

K E

earth

(Ear th Da y) = 1.08secon d s

= 30,290m /s

= 966m /s

k v

= 6

k =

8π

(1 + α) ⋅ N

𝔼

! of 44 66

Which we use to determine the radius of a proton

And is used to determine the charge of a proton

If we look at how we found that Avogadro’s number describes not just atoms, elements, and

compounds, for which it was devised in chemistry, we see it may describe all the levels of

cosmos, as we found in equations 4.5 and 4.6:

We see we can refine our equation for the radius of a proton by looking at the factors in these

equations. We can write them

Equation 9.1.

t =

2α

h 4π

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12 secon d s

2π

⋅

𝔼

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6proton s

𝔼 =

16π α

⋅ 3 2 ⋅

K E

earth

K E

moon

⋅

Ear th Da y

6.02E 23 =

16π α

⋅ 3 2 ⋅

K E

earth

K E

moon

⋅

Ear th Da y

K E

moon

K E

earth

16π α

⋅ 3 2 ⋅

⋅

Ear th Da y

16π α

= 2.67875E19

3.7331E − 20

2.225E 7

atom

! of 45 66

Where we have considered meters/kg to be the atoms in a straight line approximated very thin

then cubed to get grams per volume as a good estimate for a density of atoms in space that can

be used to study the amount of atoms in the interstellar medium to the same in a nebula to the

same in a star or planet per unit volume. We write out h and G:

We have

(

2.225E 7

atom

)

= 1.10E 22

atom s

= 0.018

= 18

h = J ⋅ s = kg

⋅ m ⋅ s =

kg ⋅ m

G = N ⋅

= kg ⋅

⋅

⋅ kg

= 6.02E 23

atom s

m ole

= kg

⋅

= kg

6.62607E − 34

6.67408E − 11

= 9.928E − 24kg

G m

9.928E − 24

1.67262E − 27

= 5935.6kg

16π α

= 2.67875E19

G m

9.928E − 24

1.67262E − 27

= 5935.6kg

16π α

⋅

G m

(

2.67875E19

)(

5935.6kg ⋅

)

1.5899885E 23secon d s ≈ 1.6E 23secon d s

= 6.02E 23

atom s

m ole

16π α

⋅

G m

6.02E 23atom s

1.6E 23secon d

= 3.7625

atom s

secon d

3 2 = 4.24264

! of 46 66

EarthDay=(24)(60)(60)=86400 seconds

We have

Equation 9.2.

That is

Equation 9.3.

Equation 9.4.

Now if we compare 9.3 with our old equation for a second we have

We see we eliminate the 16 and move into the denominator. We see that

The two are close to the same and this brings up the question; do we want the time unit be the

one closest to an actual second. The second comes from the Earth rotation dividing it into 60

units we call minutes and that into 60 units we call seconds. Certainly we see the second is

Natural, but do we want to change it a little as our fundamental unit. In terms of what we are

doing here, that depends on the exact value of the radius of a proton, and experimentally it is the

least accurate measurement we have been able to do in the laboratory to date among all our

constants used here. It depends on exactly what that is, and scientists believe they are really

(3.7525)(4.24264) = 15.962933 ≈ 16

atom s

secon d

≈ 16s

−1

(86400s)(16s

−1

) = 1,382,400a tom s

⋅

3 2π α

⋅

G m

= 1

atom

sec

3.7625

⋅ (4.24264) = 0.9976833 ≈ 1

3 2π α

= 1secon d

3 2π α

= 6E 23

= 1secon d

16π α

⋅

3 2

𝔼

⋅

= 1.12 secon d s

3 2

= 0.265165

3 2

= 0.23570

! of 47 66

honing in on it. It wavers a little depending on the method we use to measure it. But I think it is

closest to the one using just because

Thus equation 9.3 can be written

Equation 9.5.

Which fits in best with our theory of sixfold symmetry because is the diagonal of a unit

square and we then have that divided up into six equal parts.

We have said our theory is holistic and that we are describing the proton radius in terms of the

whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van der

Waals radius, which is determined by hydrogen gas, or . It is done like this:

3 2

π α

= 1secon d

! of 48 66

Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases

that behave according to kinetic-molecular theory, he described real gases which don’t. His

equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:

Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to

increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.

However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed

beyond a limit, and at close proximity to one another they interact and have a volume around

them that excludes one another, that is they have walls. He said

That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a

correction factor b. The volume of the particles is the number of particles ( ) times the volume

of one particle:

Thus there exists a sphere of radius 2r formed by two particles in contact where no other

particles can enter. It gives the correction factor

And the volume correction for n particles is

This is the volume correction to the Ideal Gas Law. The pressure correction says real gases

exhibit less pressure because their particles interact which is a net pulling by the bulk of

particles away from the container walls.

The reduction in pressure is proportional to by a factor a. We have for reduction of pressure

that

We substitute this into the Ideal Gas Law:

This can be written as a cubic

PV = n RT

= V

− b

π r

b = (4)

π r

nb = 4n ×

π r

= P

+ a

(

P + a

)

(

V − n b

)

= n RT

! of 49 66

Which allows one to compute the critical conditions of liquefaction and toderive an expression

of the principle corresponding states. In the cubic form we have as the solution three volumes

which can be used for computing the volume at and below critical temperatures.

Thus the Van Der Waals radius is estimated

For hydrogen experimentally. Therefore with

We have described the derivation of radius of a hydrogen atom from the Van Der Waals

equations that we use to get

Where the Van Der Waals equations are

So we can compute the radius of a proton from

Which needs to be multiplied through by one second to get

−

(

b +

)

V −

a b

= 0

(

π +

)

(3φ − 1) = 8τ : π =

, φ =

, τ =

π r

4π

⋅

b = 26.61

m ol

= 6.02E 23

= 1.0967E − 8cm = 1.0967E − 10m

π α

= 1secon d

(

P + a

)

(

V − n b

)

= n RT

−

(

b +

)

V −

a b

= 0

π r

4πh

= 8.288587 × 10

−16

m = 0.829f m

! of 50 66

But in determining the radius of a proton from

We need to determine k in terms of the Chandrasekhar limit where k is

Where

Where we multiply by 1.44, the Chandrasekhar limit to get 2.8634E30kg giving us our

intermediate mass using the Chandrasekhar limit as the upper limit for a mass, and

The mass of a proton for the lower limit. To have our equation k

We needed to use the Chandrasekhar limit

So we address now how that limit is derived…

2π

⋅

𝔼

4πh

= 8.288587 × 10

−16

= 0.829f m

t =

2α

h 4π

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12 secon d s

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

k =

1 + α

⋅ N

𝔼

= (2.8634E 30)(1.67262E − 27) = 69.205kg

⊙

= 1.98847E 30kg

⊙

= 1.67262E − 27kg

k =

8π

(1 + α) ⋅ N

𝔼

M ≤ 0.77

ℏ

= 1.41 ⊙

! of 51 66

The pressure of the outer shell of star balances with the outward pressure in the core of

the star (thermal pressure). Pressure is force per unit surface area thus…

is the mass of the core pulling in the mass of the shell and is the radius of the

core. The surface area of the star is that of a sphere, . We have

The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure

times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.

The number of protons in the core is . We have

Where is the Boltzmann constant ( ). Since we must have

if the star is not to implode or explode

And we have the estimate for the temperature of the core of a star.

Fusion would not occur at the low temperature of a star like the Sun in that there would not be

enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum

mechanical tunneling. The collisions are given by the kinetic energy of the particles

. We have

The velocity v yields the minimum distance between protons as the De Broglie wavelength

gravit y

P =

F = m a = PA

P =

m a

m a = G

shell

core

shell

core

A = 4π r

core

gravit y

= G

shell

core

4π r

core

ther mal

core

⋅

π r

core

1.380649E − 23J ⋅ K

−1

gravit y

= P

ther mal

core

shell

core

4πϵ

min

! of 52 66

Since the velocity is the root mean square velocity of the protons…

We have the temperature of the star is

This is another estimate. Since the mass of a star is its volume times its density

But for a star density varies with radius

If we take the derivative of both sides of the equation we have one of the equations of stellar

structure:

The so-called conservation of mass equation. The force on the shell of the star is given by the

mass of the shell

Again for there to be balance gravitation pressure equals thermal pressure:

Another equation of the equations of stellar structure. The so-called equation of hydrostatic

equilibrium. This can be written

If the star is an ideal gas the density of the star varies as where for a

monatomic gas and then

λ =

rms

min

(

4πϵ

)

3π

m =

π r

4π

∫

ρ(r)dr

dm(r)

= 4π r

ρ(r)

= − G

m(r)4π r

ρ(r)

dρ(r)

= − G

m(r)ρ(r)

(

ρ(r)

dP(r)

)

= − 4π Gr

ρr

= consta nt

γ = 5/3

! of 53 66

In stellar dynamics we write

So that

The abundance of hydrogen and helium in the universe are approximately 75% and 24%,

respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have

Ionized hydrogen and helium have and for the Sun because of high metal

content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such

pressure is given in terms of temperature (Temperature is proportional to radiation energy):

There are three kinds of pressures that can be generated by a star: gas pressure, radiation

pressure, or degeneracy pressure.

A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called

white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and

thereby collapsed under gravity but are kept from collapsing into black holes by thermal

pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a

star on the main sequence even though they are not doing fusion. It was the Indian physicist

Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity

overcome the degeneracy pressure and collapse. The non-relativistic equation is:

There are many resources available that derive this and you can find it in any textbook on

astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the

derivation to those works.

Returning to our equation for one second in terms of the Earth and Moon orbital kinetic

energies:

P ∝

∝ ρ

5/3

μm

gas

μm

4 + 12

1 + 12 + 14

= 0.59

μ = 0.59

μ = 0.62

rad

⋅

M ≤ 0.77

= 1.41 ⊙

K E

moon

K E

earth

(Ear th Da y) = 1.08secon d s

! of 54 66

We notice it would be exactly one second if the Earth day were shorter. And, indeed it was a long

time ago.

For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to

the moon. The days become longer by 0.0067 hours per million years. Our Equation

Is actually 1.2 seconds for KE of moon and KE or earth calculated with their average orbital

velocities. We will compute how long ago the Earth day was what it was needed to turn that 1.2

seconds into one second, and then how long ago the Earth day was what it was needed to turn

the 1.08 seconds into one second.

We have

24-20=0.0067t

t=597 million years

This was when the earth went through a dramatic change and there was a big explosion of life

(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went

extinct 65 million years ago giving small mammals a chance to evolve paving the way for

humans.

24-x=0.0067t

x=23.5645 hours

We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic

calendar:

What is the next term?

20+3+0.57735+0.4714=24hours

Which bring us to today. Now the calculation for the 1.08 seconds that works with KE of moon

and KE earth calculated using aphelions and perihelions.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

24h ours

1.2

= 20hours

3cos(0

∘

) +

cos(30

∘

) = din osaur − ex t inct ion =

20hrs + 3hrs +

hrs +

24h ours

1.08

= 22.222h ours

! of 55 66

24-22.222=0.0067t

t=265.373 million years

This is the Permian period which was from 299 to 251 million years ago, it was at the end of the

Paleozoic Era which was followed by the Mesozoic Era. The distinction between the Paleozoic

and the Mesozoic is made at the end of the Permian to mark the largest mass extinction

recorded in Earth’s history.

Thus, to conclude, we write

And focus for a bit on . It takes the square which has diagonal to side is and divides that

up with regular hexagon which is six radii and six sides equation one another reconciling Euler’s

number e with golden ratio characteristic of fivefold symmetry, the regular pentagon. That is

Equation 9.6.

Now we show what this is geometrically and explore how we arrived at it in the following

pages…

π α

= 1secon d

Φ(1 + Φ)

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Appendix 1

We suggested the second was a natural unit, and that, if it was, should be in the orbital

mechanics of the earth because the second comes from the calendar, which is based on the

orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the

rotation of the Earth. We found that it was as the following:

Let’s show that…

To find the translational kinetic energy of the moon:

Distance from earth: 3.85E8m

Orbital period:

T=27.32 days=2.36E6seconds

v=1.025E3m/s~1000m/s

Mass: 7.34767E22kg

Use

E=3.67E28 Joules

To find the translational kinetic energy of the earth:

Distance from Sun: 1AU=1.496E11m

Orbital period: 1 year=3.1558E7 seconds

v=2.9785E4m/s, Earth mass: 5.9722E24kg

E=2.649E33 Joules

Earth day=(24)(60)(60)=86,400 seconds.!

moon

earth

(EarthDay) ≈ 1second

K . E . Moon = 3.67E 28J

K . E . Ear th = 2.649E 33j

2π (3.85E8m) = 2.419E 9m

E =

2π (1.496E11m) = 9.399E11m

! of 64 66

Appendix 2

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number. We

make a program that looks for close to whole number solutions so we can create a table of values

for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. Here is the code for the program:!

h4π r

! of 65 66

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?):

");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}!

! of 66 66

The Author

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